Math, asked by skazzu, 11 months ago

The taxi charges in Hyderabad are fixed, along with the charge for the distance covered.
Upto first 3 km you will be charged a certain minimum amount. From there onwards you
have to pay additionally for every kilometer travelled. For the first 10 km, the charge paid
is 166. For a journey of 15 km. the charge paid is 256.
i What are the fixed charges and charge per km?
ii How much does a person have to pay for travelling a distance of 25 km?​

Answers

Answered by ShreyaSingh31
62

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Given :-

  • Charge for first 10 km is ₹166
  • For a journey of 15 km charge paid is ₹256

To find :-

  • the fixed charges
  • charge per km
  • Amount to be payed for travelling a distance of 25 km

Solution :-

We will first calculate the fixed charges and the charges per km.

As per the first condition,

  • For the first 10 km, the charge paid is 166,

So, let the fixed charge be x

Let the charge per km be y

For first 10 km charge paid is ₹ 166.

Let's represent it mathematically to get our first equation,

x + 10y = 166 -----> 1

[ x is fixed charge, which means it constant. We can't make any changes in x, so thereby, as per assumption, y is charge per km, which can vary so we can make changes in it]

As per the second condition,

  • For a journey of 15 km. the charge paid is 256

Let's represent this too mathematically to get our second equation,

x + 15y = 256 ----> 2

We got our two equations, solve them via elimination method, simultaneously.

Subtract equation 2 from 1,

...x + 10y = 166

- (x + 15y = 256)

-----------------------

- 5y = -90

5y = 90

y = \bf\large\frac{90}{5}

y = 18

•°• Charge per km = y = 18

Substitute, y = 18 in equation 1,

x + 10y = 166

x + 10 (18) = 166

x + 180 = 166

x = 166 - 180

x = - 14

As, x = charge (fixed) it cannot be negative,

•°• x = 14

° Fixed charge = x = 14

Now, we have to find the charge payed for a journey of 25 km

We will add the fixed charge by 25 times the charge per km,

For 25 km,

\bf\implies x + 25y

\bf\implies 14 + 25(18)

\bf\implies 14 + 450 = 464

° Charge payed for a journey of 25 km is 464

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